Solar System Voltage Drop Calculations in Excel

[Calif.C10]

I have been tasked to work on the design for a fairly small PV project but it has some distances involved that have everyone concerned. I am wondering if anyone would have an Excel program for doing voltage drop calculations they would be willing to share. Or, if they have the correct instructions on how to properly write the formula(s) in each cell? I need to OLE insert the VD calculations into the CAD drawing(s).

Any help on this would be greatly appreciated.

Thanks.

 

[ron]

I searched google for a xls file with a search term volt drop
http://www.oynot.com/media/Basic_Voltage_drop_calculator_.xls
maybe there are others too.

 

[gar]

110510-1648 EDT

Cal.C10:

First questions are about how this PV system is structured.
1. Are all the PV panels tied together at the DC level at the array?
If so then how large is the array in length and width in feet?
2. Are micro-inverters being used?
If not, then why not?
3. What is the distance from the array to where the power is used, stored, or whatever?
4. Does the power have to remain DC?
5. What is maximum voltage and current level?
6. Are you concerned about the voltage drop within the array area?

What is your background in electrical circuit theory?

Suppose your concern is send DC over long lines, say 5000 ft between source ad destination. Consider 300 V maximum at the array, and a maximum current of 100 A. To hold the voltage drop to 30 V @100 A the total loop resistance, 10,000 ft, has to be less than 30/100 = 0.3 ohms, 0.03 ohms/1000 ft.

This would require ridiculously large copper wire. So change the DC to AC at the array, and then step up to 3000 V with a transformer. Retain the same voltage drop, 30 V but at the 3000 V level. Now the current is only 10 A, 30/10 = 3 ohms, or 0.3 ohms/1000 ft. This would get you to about #4 wire. Also consider the original 10% drop instead of 1%. Now the wire size could drop to about #14.

These are just to illustrate alternatives.

You need a clear definition of many parts of the problem.

Just a formula to plug values into will not necessarily provide a solution to a problem. In fact as you work on a problem the definition of the problem may change. The problem might be what is the most cost effective means to transfer power from the array to its destination. This would include capital cost, operational cost, and the value derived from the product.

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[Calif.C10]

Sheesh....

Sheesh....

See Remarks:

110510-1648 EDT

Cal.C10:

First questions are about how this PV system is structured.
1. Are all the PV panels tied together at the DC level at the array?Yes-Strings of 14 Modules
If so then how large is the array in length and width in feet? It varies- they are all 26' in width and either 60', 76' or 96' in length.
2. Are micro-inverters being used?No-Array Strings to Combiners to DC Disconnect to Inverter-Inverter to AC Disconnect to Line side Bussing Tap ahead of the MCB
If not, then why not? Huh? One Main Inverter
3. What is the distance from the array to where the power is used, stored, or whatever? Approximately 350 Feet-String Wiring to Combiner-this is the worst case scenario
4. Does the power have to remain DC? See response to #2
5. What is maximum voltage and current level? 515.2 Volts DC. (Voc) [V] from 14 modules in one string. 7.68 amps @ Pmax (Imp) [A]
6. Are you concerned about the voltage drop within the array area? Nope!

What is your background in electrical circuit theory? Why are you busting my chops? It was a fairly simple question. I have found MANY answers (not all here in this website) that didn't require the follow up you have asked. My goodness, I input some simple data in my $60 Electri-Calc calulator, and it spits out an answer-not more questions...

Suppose your concern (which it is not) is send DC over long lines, say 5000 ft between source and destination. Consider 300 V maximum at the array, and a maximum current of 100 A. To hold the voltage drop to 30 V @100 A the total loop resistance, 10,000 ft, has to be less than 30/100 = 0.3 ohms, 0.03 ohms/1000 ft.

This would require ridiculously large copper wire. So change the DC to AC at the array, and then step up to 3000 V with a transformer. Retain the same voltage drop, 30 V but at the 3000 V level. Now the current is only 10 A, 30/10 = 3 ohms, or 0.3 ohms/1000 ft. This would get you to about #4 wire. Also consider the original 10% drop instead of 1%. Now the wire size could drop to about #14.

These are just to illustrate alternatives.

You need a clear definition of many parts of the problem. What problem? I just wanted an Excel file...

Just a formula to plug values into will not necessarily provide a solution to a problem.(Yes it does in my case) In fact as you work on a problem the definition of the problem may change. The problem might be what is the most cost effective means to transfer power from the array to its destination. This would include capital cost, operational cost, and the value derived from the product.

Basically I was looking for a formula which I can insert via OLE from Excel into my CAD drawing so I can prove I can use #12 USE-2 CU for my string wiring-that's all....

I knew there was a reason I strayed away from this web site for as many years as I did.

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[gar]

110511-0759 EDT

Calif.C10:

I think your original question should have been something like:
What is an approximate equation for the voltage drop in a copper wire circuit where the following are given as inputs:
Total wire length is in feet,
DC current in amps, and
Wire area in circular mills?

At room temperature, 20 C, the resistance of annealed copper wire is
R = K*L/CM where K = 10.37, L length in ft, CM is circular mills
then
Vdrop = I*R = I*K*L/CM

K is different for other materials and temperatures.

Since a circuit has two wires the constant K can be doubled and the L variable can be the distance from input to output. Then the wire temperature rise needs to be considered.

NEC simplifies this to a constant of 25. But for long runs where the current density is lower because the primary criteria is voltage drop and not current capacity, then a more accurate constant is between 20.74 and 25.

The NEC equation is apparently
Vdrop = 25*I*L/CM.

From
http://en.wikipedia.org/wiki/Circular_mil
you can get the formula to convert wire size to CM.

AWG circular mil formula

The formula to calculate the circular mil for any given AWG (American Wire Gauge) size is as follows. An represents the circular mil area for the AWG size n.

equation should be here

For example, a number 12 gauge wire would use n = 12; and the calculated result would be 6529.946789 circular mils
Sizes with multiple zeros are successively larger than the number 0 gauge size and can be denoted using "number of zeros/0"; for example 4/0 for the number 0000 gauge. For an m/0 AWG wire size, use n = −(m−1) = 1−m in the above formula.

For example, the number 0000 gauge or 4/0 gauge, would use n = −3; and the calculated result would be 211,600 circular mils

The equation does not display in the quote so refer to the Wikipedia site.

When someone just asks for an equation this is an immediate red flag. Without an understanding of the basics, and some idea of how an equation is derived, the arbitrary use of an equation can easily lead to bad results in the use of the equation.

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